| 141 Experiment 9: Serial and Paralilel Dilutions 11 Section Date Name POST LAB

Question

| 141 Experiment 9: Serial and Paralilel Dilutions 11 Section Date Name POST LAB QUESTIONS Cupric sulfate has a molar absorptivity of 2.81 M-1 cm-1 at max-635nm. 1. A cupric sulfate stock solution was analyzed in a spectrometer with a 1.0 em pathlength cuvette. a. (2 pts) What is the molarity of a cupric sulfate stock solution if the absorbance was o b. (3 pts) What is the %T for the stock solution? c. (3 pts) What is the concentration of a second solution made by diluting 15 .00 mL of the stock solution in a 25 mL volumetric flask? d (2 pts) What is the absorbance and %T of the diluted solution? e. (2 pts) What is the absorbance of the diluted solution if the wavelength of the spectrophotometer was changed to 535 nm? f. (5 pts) Fill in the table below using data calculated above Solution Stock Diluted %T 0.880 e. (3 pts) Write the Beer-Lambert equation for the data in the table above. As 281M-1cm-1 [Cuso,4] + 0.0

Explanation / Answer

a)

from Beer’s Law

A = e*l*C

A = absorbance of sample

e is the molar absorptivity , the typical units are 1/M-cm

l size of cuvette, typically reported in cm

c is the molar concentration, in mol per ltier or M

apply

0.88 = 2.81*1*C

C = 0.88/2.81

C = 0.31316 mol per liter

b)

find %T

A = 2 - log(T%)

0.88 = 2- log(T%)

T% = 10^-(0.88-2)

T% = 13.18256

c)

when diluting 15 mL to 25 mL

We need to apply dilution law, which is based on the mass conservation principle

initial mass = final mass

this apply for moles as weel ( if there is no reaction, which is the case )

mol of A initially = mol of A finally

or, for this case

moles of A in stock = moles of A in diluted solution

Recall that

mol of A = Molarity of A * Volume of A

then

moles of A in stock = moles of A in diluted solution

Molarity of A in stock * Volume of A in stock = Molarity of A in diluted solution* Volume of A in diluted solution

Now, substitute known data

M1*V1 = M2*V2

0.31316 *15 = M2*25

M2 = 0.31316 *15 /25 = 0.187896 mol per liter

d)

New A/T%

A = e*l*C

A = 2.81*1*0.187896

A = 0.5279

A = 2-log(T%)

0.5279 = 2-log(T%)

T% = 10^-(0.5279-2)

T% = 29.655

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